2017 ACM-ICPC 亚洲区（西安赛区）网络赛： B. Coin 【概率题】【数论】

Bob has a not even coin（就是一个不均匀的硬币，朝上的概率不一定是1/2）, every time he tosses the coin, the probability that the coin's front face up is q/p(q/p<=1/2).

The question is, when Bob tosses the coin k times, what's the probability that the frequency of the coin facing up is even number（就是硬币朝上次数为偶数次的概率和，0次也算）.

If the answer is X/Y​​, because the answer could be extremely large, you only need to print (X * Y^{-1}) /mod (10^9+7).

Input Format

First line an integer TT, indicates the number of test cases (T≤100).

Then Each line has 33 integer p,q,k(1<= p,q,k ,k<=10^7) indicates the i-th test case.

Output Format

For each test case, print an integer in a single line indicates the answer.

样例输入

2
2 1 1
3 1 2

样例输出

500000004
555555560

思路:

题目说的有点晦涩。

一个组合数学题。。。。。

　

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 using namespace std;
 const int MOD=1e9+;
 typedef long long ll;
 ll p,q,k;

 ll qmod(ll x,ll n)
 {
     ll res=;
     while(n){
         if(n&) res=(res*x)%MOD;
         x=(x*x)%MOD;
         n/=;
     }
     return res;
 }

 int main()
 {
     int T;
     cin>>T;
     while(T--)
     {
         cin>>p>>q>>k;
         ll x=qmod(p,k);
         ll y=qmod(p-*q,k);
         ll ans=((x+y)%MOD*qmod(*x,MOD-))%MOD;
         cout<<ans<<endl;
     }
     return ;
 }